Question

please give the answer of my question.... mathematics fast​

Answer 1

Step-by-step explanation:

$\rm {\huge\star} \: Given$

$\bigg(\rm\dfrac{1}{1 + x^{a - b}} \bigg) + \bigg( \dfrac{1}{1 + x ^{b - a} } \bigg) = 1$

In any number when you get a negative power it means you have to make it positive by twisting it or it means you have to divide. In this case , we have to divide. -B means ÷ B So, let's see!!

$\rm \longmapsto \bigg( \dfrac{1}{1 + \dfrac{x ^{a} }{x ^{b} } } \bigg) + \bigg( \dfrac{1}{1 + \dfrac{x^{b} }{x ^{a} } } \bigg) = 1$

Now , the value which is at down will be getting multiplied by 1 also which is at down will remain as denominator.

$\rm \longmapsto \bigg( \dfrac{1}{ \dfrac{x ^{a} + x ^{b} }{x ^{b} } } \bigg) + \bigg( \dfrac{1}{ \dfrac{x^{b} + {x}^{a} }{x ^{a} } } \bigg) = 1$

Now the denominator will go to up as soon as we do reciprocal of it .

So, it becomes

$\rm{ \tt \leadsto} \dfrac{ {x}^{b} }{ {x}^{a} + {x}^{b} } + \dfrac{ {x}^{a} }{ {x} ^{b } + {x}^{a} } = 1$

Now, the denominator are same because ; a+b = b+a

So, let's solve it!!

$\rm{ \tt \leadsto} \cancel \dfrac{ {x}^{b} + {x}^{a} }{ {x}^{a} + {x}^{b} } \longmapsto1=1$

$\rm {\huge \therefore} \: On \: simplifying \: the \: value \: we \: got \: the \: answer \: \bf1$

Hence proved

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Answer 2

Question

Prove that -: $\sf \dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1$

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Answer

We'll first simplify the given problem and later compare if the answer obtained after simplifying is the same as the one required.

Using this law of exponents, let's expand the problem → $\sf \frac{a^{m}}{a^{n}} = a^{m-n}$

$\sf \implies \dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}$

$\sf \implies \dfrac{1}{1+\dfrac{x^{a}}{x^{b}}}+\dfrac{1}{1+\dfrac{x^{b}}{x^{a}}}$

Now let's simplify the denominators.

$\sf \implies \dfrac{1}{1+\dfrac{x^{a}}{x^{b}}}+\dfrac{1}{1+\dfrac{x^{b}}{x^{a}}}$

$\sf \implies \dfrac{1}{\dfrac{1 \times x^{b} }{1\times x^{b} } +\dfrac{x^{a}}{x^{b}}}+\dfrac{1 }{\dfrac{1\times x^{a} }{1\times x^{a} } +\dfrac{x^{b}}{x^{a}}}$

$\sf \implies \dfrac{1}{\dfrac{x^{b} }{x^{b} } +\dfrac{x^{a}}{x^{b}}}+\dfrac{1 }{\dfrac{x^{a} }{x^{a} } +\dfrac{x^{b}}{x^{a}}}$

$\sf \implies \dfrac{1}{\dfrac{x^{b}+x^{a}}{x^{b}} } + \dfrac{1}{\dfrac{x^{a}+x^{b}}{x^{a}} }$

Now let's further simplify the problem.

$\sf \implies 1 \div \dfrac{x^{b}+x^{a}}{x^{b}}+1\div \dfrac{x^{a}+x^{b}}{x^{a}}$

$\sf \implies 1 \times \dfrac{x^{b}}{x^{b}+x^{a}} + 1\times \dfrac{x^{a}}{x^{a}+x^{b}}$

$\sf \implies \dfrac{x^{b}}{x^{b}+x^{a}} + \dfrac{x^{a}}{x^{a}+x^{b}}$

$\sf \implies \dfrac{x^{b}+x^{a}}{x^{b}+x^{a}}$

$\sf \implies \dfrac{1}{1}$

$\sf \implies 1$

Hence proved that →  $\sf \bf \dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1$

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